Hello Students,

Trigonometry formulas for Class 10 are provided here for students. Trigonometry is the study of relationships between angles, lengths, and heights of triangles. It includes ratios, functions, identities, and formulas to solve problems based on it, especially for right-angled triangles. Applications of trigonometry are also found in engineering, astronomy, Physics and architectural design. This chapter is very important as it comprises many topics like Linear Algebra, Calculus and Statistics.

Trigonometry is introduced in CBSE Class 10. It is a completely new and tricky chapter where one needs to learn all the formulas and apply them accordingly. Trigonometry Class 10 formulas are tabulated below.

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### List of Trigonometric Formulas for 10th Class

All the important formulas introduced to students in Class 10 are available here. Students can learn these formulas anytime from here and solve trigonometry related problems.

The trigonometric formulas for ratios are majorly based on the three sides of a right-angled triangle, such as the adjacent side or base, perpendicular and hypotenuse (See the above figure).  Applying Pythagoras theorem for the given right-angled triangle, we have:

(Perpendicular)+ (Base)= (Hypotenuse)2

⇒ (P)+ (B)= (H)2

Now, let us see the formulas based on trigonometric ratios (sine, cosine, tangent, secant, cosecant and cotangent)

### Basic Trigonometric formulas

The Trigonometric formulas are given below:

 S.no Property Mathematical value 1 sin A Perpendicular/Hypotenuse 2 cos A Base/Hypotenuse 3 tan A Perpendicular/Base 4 cot A Base/Perpendicular 5 cosec A Hypotenuse/Perpendicular 6 sec A Hypotenuse/Base

### Reciprocal Relation Between Trigonometric Ratios

 S.no Identity Relation 1 tan A sin A/cos A 2 cot A cos A/sin A 3 cosec A 1/sin A 4 sec A 1/cos A

### Trigonometric Sign Functions

• sin (-θ) = − sin θ
• cos (−θ) = cos θ
• tan (−θ) = − tan θ
• cosec (−θ) = − cosec θ
• sec (−θ) = sec θ
• cot (−θ) = − cot θ

### Trigonometric Identities

1. sin2A + cos2A = 1
2. tan2A + 1 = sec2A
3. cot2A + 1 = cosec2A

### Periodic Identities

• sin(2nπ + θ ) = sin θ
• cos(2nπ + θ ) = cos θ
• tan(2nπ + θ ) = tan θ
• cot(2nπ + θ ) = cot θ
• sec(2nπ + θ ) = sec θ
• cosec(2nπ + θ ) = cosec θ

### Complementary Ratios

• sin(π/2 − θ) = cos θ
• cos(π/2 − θ) = sin θ
• tan(π/2 − θ) = cot θ
• cot(π/2 − θ) = tan θ
• sec(π/2 − θ) = cosec θ
• cosec(π/2 − θ) = sec θ

• sin(π − θ) = sin θ
• cos(π − θ) = -cos θ
• tan(π − θ) = -tan θ
• cot(π − θ) = – cot θ
• sec(π − θ) = -sec θ
• cosec(π − θ) = cosec θ

• sin(π + θ) = – sin θ
• cos(π + θ) = – cos θ
• tan(π + θ) = tan θ
• cot(π + θ) = cot θ
• sec(π + θ) = -sec θ
• cosec(π + θ) = -cosec θ

• sin(2π − θ) = – sin θ
• cos(2π − θ) = cos θ
• tan(2π − θ) = – tan θ
• cot(2π − θ) = – cot θ
• sec(2π − θ) = sec θ
• cosec(2π − θ) = -cosec θ

### Sum and Difference of Two Angles

• sin (A + B) = sin A cos B + cos A sin B
• sin (A − B) = sin A cos B – cos A sin B
• cos (A + B) = cos A cos B – sin A sin B
• cos (A – B) = cos A cos B + sin A sin B
• tan(B) = [(tan tan B)/(– tan tan B)]
• tan(A – B) = [(tan A – tan B)/(1 + tan tan B)]

### Double Angle Formulas

• sin 2A = 2 sin A cos A = [2 tan A /(1 + tan2A)]
• cos 2A = cos2A – sin2A = 1 – 2 sin2A = 2 cos2A – 1 = [(1 – tan2A)/(1 + tan2A)]
• tan 2A = (2 tan A)/(1 – tan2A)

### Triple Angle Formulas

• sin 3A = 3 sinA – 4 sin3A
• cos 3A = 4 cos3A – 3 cos A
• tan 3A = [3 tan A – tan3A]/[1 − 3 tan2A]

## Solved Examples

Example 1:

If sin A = ⅗, then find the value of cos A and cot A.

Solution:

Given,

sin A = ⅗

Using the identity, sin2A + cos2A = 1,

cos2A = 1 – (⅗)2

= (25 – 9)/25

= 16/25

cos A = ⅘

Also, cot A = cos A/sin A = (⅘)/(⅗) = 4/3

Example 2:

Evaluate sin 35° cos 55° + cos 35° sin 55°.

Solution:

Given expression:

sin 35° cos 55° + cos 35° sin 55°

This is of the form sin A cos B + cos A sin B.

Thus, by using the identity sin(A + B) = sin A cos B + cos A sin B, we get;

sin 35° cos 55° + cos 35° sin 55° = sin(35° + 55°) = sin 90° = 1

Example 3:

If tan P = cot Q, then prove that P + Q = 90°.

Solution:

Given,

tan P = cot Q

As we know, tan(90° – A) = cot A.

So, tan P = tan(90° – Q)

Therefore, P = 90° – Q

And

P + Q = 90°

Hence proved.